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GMATinsight
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This is my approach:

As there are 3 walls there can be 3 places for portraits on each. So (3+3+3=9) 9C3=24

Hope this will help
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GMATinsight
In how many ways can 3 different portraits be mounted on 3 of the four walls inside a room???

A) 4
B) 6
C) 12
D) 16
E) 24

Source: https://www.GMATinsight.com

The number of ways to pick 3 walls from 4 is 4C3 = 4. Once 3 walls are picked, the number of ways to mount the portraits onto the walls is 3! = 6. Therefore, there are a total of 4 x 6 = 24 ways to mount 3 portraits on 3 of the four walls inside a room.

Answer: E
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In how many ways can 3 different portraits be mounted on 3 of the four walls inside a room???

A) 4
B) 6
C) 12
D) 16
E) 24

Source: https://www.GMATinsight.com

The number of ways to pick 3 walls from 4 is 4C3 = 4. Once 3 walls are picked, the number of ways to mount the portraits onto the walls is 3! = 6. Therefore, there are a total of 4 x 6 = 24 ways to mount 3 portraits on 3 of the four walls inside a room.

Answer: E

Hi can someone help on this one? Why is it that 3 different portraits can be mounted on 3 walls 3! ways?
How come it's not 6! / 3! x 3! = 20 ways?

ppp ||| <---- 3 portraits and 3 divisors so 6!/3! x 3!

I guess the idea here is that there can be a maximum of one portrait per wall...
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GMATinsight
In how many ways can 3 different portraits be mounted on 3 of the four walls inside a room???

A) 4
B) 6
C) 12
D) 16
E) 24

Source: https://www.GMATinsight.com

The number of ways to pick 3 walls from 4 is 4C3 = 4. Once 3 walls are picked, the number of ways to mount the portraits onto the walls is 3! = 6. Therefore, there are a total of 4 x 6 = 24 ways to mount 3 portraits on 3 of the four walls inside a room.

Answer: E

Hi can someone help on this one? Why is it that 3 different portraits can be mounted on 3 walls 3! ways?
How come it's not 6! / 3! x 3! = 20 ways?

ppp ||| <---- 3 portraits and 3 divisors so 6!/3! x 3!

I guess the idea here is that there can be a maximum of one portrait per wall...

Response:

As a matter of fact, you are correct; the question should have said “one painting per wall.” However, your answer of 20 is not the correct answer for that scenario (unless the paintings are identical, which we know is not the case). If we can mount more than one painting per wall, the number of ways should increase. If that was the case, you would solve “prq |||” instead of “ppp |||”.
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Hi everyone,

In case more than one painting could have been hung on each wall, what would the solution have been?
Four alternatives for each painting, right?
So 4^3, correct?

And what about if also the order (let's say from left to right, or from bottom to top) of two or more paintings on the same wall would have mattered?
4 * 5 * 6 = 120
Is it correct?

Thanks

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